Water coming out of a horizontal tube at a speed $v$ strikes a vertical wall normally,close to the mouth of the tube,and falls down vertically after impact. When the speed of water is increased to $2v$:

$A$ liquid flows through a horizontal tube. The velocities of the liquid in the two sections,which have areas of cross-section $A_1$ and $A_2$,are $v_1$ and $v_2$ respectively. The difference in the levels of the liquid in the two vertical tubes is $h$.

$A$ cylindrical tube,with its base as shown in the figure,is filled with water. It is moving down with a constant acceleration $a$ along a fixed inclined plane with angle $\theta=45^{\circ}$. $P_1$ and $P_2$ are pressures at points $1$ and $2$,respectively,located at the base of the tube. Let $\beta=(P_1-P_2) / (\rho g d)$,where $\rho$ is the density of water,$d$ is the inner diameter of the tube,and $g$ is the acceleration due to gravity. Which of the following statement$(s)$ is(are) correct?
$(A)$ $\beta=0$ when $a=g / \sqrt{2}$
$(B)$ $\beta>0$ when $a=g / \sqrt{2}$
$(C)$ $\beta=\frac{\sqrt{2}-1}{\sqrt{2}}$ when $a=g / 2$
$(D)$ $\beta=\frac{1}{\sqrt{2}}$ when $a=g / 2$

We are able to squeeze snow and make balls out of it because of

$A$ $U$-tube has a left limb with a cross-sectional area one-quarter that of the right limb. The tube contains mercury (density $13.6 \text{ g/cm}^3$). The level of mercury in the narrow (left) limb is at a distance of $36 \text{ cm}$ from the upper end of the tube. What will be the rise in the level of mercury in the right limb if the left limb is filled to the top with water (in $\text{ cm}$)? (Density of water = $1 \text{ g/cm}^3$)

The vertical limbs of a $U$ shaped tube are filled with a liquid of density $\rho$ up to a height $h$ on each side. The horizontal portion of the $U$ tube having length $2h$ contains a liquid of density $2\rho$. The $U$ tube is moved horizontally with an acceleration $g/2$ parallel to the horizontal arm. The difference in heights in liquid levels in the two vertical limbs,at steady state,will be: